Half life id independent of the concentration of `A`. After `10 mi n` volume of `N_(2)` gas is `10 L` and after complete reaction is `50 L`. Hence, the rate constant is

Options

A. `(2.303)/(10) log 5 mi n^(-1)`
B. `(2.303)/(10) log 1.25 mi n^(-1)`
C. `(2.303)/(10) log 2 mi n^(-1)`
D. `(2.303)/(10) log 4 mi n^(-1)`

Answer

B

Explanation

It follows first order kinetics Since `T_(50)` is independent of concentration.

Hence, `x=10L, a= 50L`
`:. k=(2.303)/(10) log.(50)/(40)=(2.303)/(10)log 1.25 min^(-1)`